Question

Cho bt B=\(\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

Answer 1

ĐKXĐ:x\(\ge0\)

B=\(\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{\sqrt{x}+1}{x-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

=\(\left(\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

=\(\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\sqrt{x}}=\dfrac{2}{\sqrt{x}+1}\)

Answer 2

điều kiện \(x>0;x\ne1;\)

\(B=\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

\(B=\left(\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

\(B=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(B=\dfrac{2}{\sqrt{x}+1}\)